Chapter 15, Problem 15.106SP

a)

Interpretation Introduction

Interpretation:

The given equation is the prediction based on Le-Chatelier’s principle about the shift in equilibrium with temperature has to be interpreted.

Concept introduction:

Law of mass action: The rate of chemical reaction is directly proportional to the product of concentrations of reactant to products.

$\begin{array}{l}\text{aA}\text{+}\text{bB}\to \text{cC}\text{+}\text{dD}\\ \\ {\text{K}}_{eq}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}\text{for}\text{aqueous}\\ \\ {\text{K}}_{\text{eq}}\text{=}\frac{{\left({\text{P}}_{\text{C}}\right)}^{\text{c}}{\left({\text{P}}_{\text{D}}\right)}^{\text{d}}}{{\left({\text{P}}_{\text{A}}\right)}^{\text{a}}{\left({\text{P}}_{\text{B}}\right)}^{\text{b}}}\text{for}\text{gases}\\ \end{array}$

Van’t Hoff equation: the change in equilibrium constant, ${\text{K}}_{eq}$ of a chemical reaction to the change in temperature, T given the standard enthalpy change $\Delta H^{\circ }$ for the process.

Clausius-Claypeyron equation:

$\begin{array}{l}\text{lnP}\text{=}\frac{{\text{-ΔH}}^{\text{o}}}{\text{RT}}\\ where,\\ \text{P}\text{varpor}\text{pressure}\text{atTemperature}\text{T}\left(\text{in}\text{K}\right)\\ {\text{ΔH}}^{\text{o}}\text{enthalpy}\text{of}\text{vaporization}\text{for}\text{the}\text{substance}\\ \text{R}\text{=}\text{8}\text{.314}\text{J/mol}\text{.K}\end{array}$

Explanation of Solution

Let’s write the Van’t Hoff equation at two different temperatures:

At ${\text{T}}_{\text{1}}$:  $\ln K_1=\frac{-\Delta H^{\circ }}{R{T}_1}+C$------ (1)

At ${\text{T}}_2$:  $\ln K_2=\frac{-\Delta H^{\circ }}{R{T}_2}+C$------ (2)

Taking the difference between two equations,

$\begin{array}{l}\ln K_1-\ln K_2=\frac{-\Delta H^{\circ }}{R{T}_1}+C-\left(\frac{-\Delta H^{\circ }}{R{T}_2}+C\right)\\ \\ =\frac{-\Delta H^{\circ }}{R{T}_1}+\frac{\Delta H^{\circ }}{R{T}_2}\\ \\ \ln \left(\frac{K_1}{K_2}\right)=\frac{\Delta H^{\circ }}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\end{array}$

Assuming an endothermic reaction, $\Delta H^{\circ }>0$ and ${\text{T}}_{\text{2}}{\text{>T}}_{\text{1}}$. Then, $\frac{\Delta H^{\circ }}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)<0$ meaning that

$\ln \left(\frac{K_1}{K_2}\right)<0$ or ${\text{K}}_{\text{1}}\text{<}{\text{K}}_{\text{2}}$. A larger ${\text{K}}_{\text{2}}$ indicates that there are more products at equilibrium as the temperature is raised. This agrees with Le-Chatelier’s principle that an increase in temperature favors the forward endothermic reaction. The opposite of the above discussion holds for an endothermic reaction.

b)

Interpretation Introduction

Interpretation:

The molar heat of vaporization of water has to be calculated.

Concept introduction:

Law of mass action: The rate of chemical reaction is directly proportional to the product of concentrations of reactant to products.

$\begin{array}{l}\text{aA}\text{+}\text{bB}\to \text{cC}\text{+}\text{dD}\\ \\ {\text{K}}_{eq}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}\text{for}\text{aqueous}\\ \\ {\text{K}}_{\text{eq}}\text{=}\frac{{\left({\text{P}}_{\text{C}}\right)}^{\text{c}}{\left({\text{P}}_{\text{D}}\right)}^{\text{d}}}{{\left({\text{P}}_{\text{A}}\right)}^{\text{a}}{\left({\text{P}}_{\text{B}}\right)}^{\text{b}}}\text{for}\text{gases}\\ \end{array}$

Multiple equilibria: If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

$\begin{array}{l}\underset{\_}{\begin{array}{l}\text{A}\text{+ B }\rightleftarrows {\text{ C + D K}}_{\text{c}}^{\text{'}}\\ \text{C}\text{+ D }\rightleftarrows {\text{ E + F K}}_{\text{c}}^{\text{''}}\end{array}}\\ \text{overall}\text{reaction:}\underset{\_}{\text{A}\text{+ B }\rightleftarrows \text{ E + F}{\text{K}}_{\text{c}}}\end{array}$

The equilibrium constant for two separate equilibrium constants are,

${\text{K}}_{\text{c}}^{\text{'}}\text{=}\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]\left[\text{B}\right]}\text{and}{\text{K}}_{\text{c}}^{\text{''}}\text{=}\frac{\left[\text{E}\right]\left[\text{F}\right]}{\left[\text{C}\right]\left[\text{D}\right]}$

For overall reaction, the equilibrium constant ${\text{K}}_{\text{c}}$ is,

${\text{K}}_{\text{c}}^{\text{'}}{\text{K}}_{\text{c}}^{\text{''}}\text{=}\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]\left[\text{B}\right]}\times \frac{\left[\text{E}\right]\left[\text{F}\right]}{\left[\text{C}\right]\left[\text{D}\right]}=\frac{\left[\text{E}\right]\left[\text{F}\right]}{\left[\text{A}\right]\left[\text{B}\right]}$

Therefore, $\boxed{{\text{K}}_c\text{=}{\text{K}}_{\text{c}}^{\text{'}}\times {\text{K}}_{\text{c}}^{\text{''}}}$

Van’t Hoff equation: the change in equilibrium constant, ${\text{K}}_{eq}$ of a chemical reaction to the change in temperature, T given the standard enthalpy change $\Delta H^{\circ }$ for the process.

$\ln K_{eq}=\frac{-\Delta H^{\circ }}{RT}+C$

Clausius-Claypeyron equation:

$\begin{array}{l}\text{lnP}\text{=}\frac{{\text{-ΔH}}^{\text{o}}}{\text{RT}}\\ where,\\ \text{P}\text{varpor}\text{pressure}\text{atTemperature}\text{T}\left(\text{in}\text{K}\right)\\ {\text{ΔH}}^{\text{o}}\text{enthalpy}\text{of}\text{vaporization}\text{for}\text{the}\text{substance}\\ \text{R}\text{=}\text{8}\text{.314}\text{J/mol}\text{.K}\end{array}$

Explanation of Solution

For the given reaction,

     ${\text{H}}_2\text{O }\left(\text{l}\right)\stackrel{}{\rightleftharpoons }{\text{H}}_2\text{O}\left(\text{g}\right)\Delta H_{vap}=?$

The given reaction is heterogeneous equilibrium, so, the equilibrium constant expression for gas is,

Rearranging ${\text{K}}_{\text{P}}$ expression:  $K_P=P_{H_{2}O}$ as $P_{H_{2}O}$ of standard state of water (${\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\left(\text{liquid}\right)$) is one.

Given: $\begin{array}{l}{T}_1={30}^{\circ }C+273=303K;K_{H_{2}O}=31.82mmHg\\ T_2=\text{ 5}{0}^{\circ }C+273=323K;K_{H_{2}O}=92.51mmHg\end{array}$

Substituting given value into the derived Van’t Hoff equation as,

$\begin{array}{l}\text{ln}\left(\frac{{\text{K}}_{\text{1}}}{{\text{K}}_{\text{2}}}\right)\text{=}\frac{{\text{ΔH}}^{\text{o}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right)\\ \\ \text{ln}\left(\frac{\text{31}\text{.82}\text{mmHg}}{\text{92}\text{.51}\text{mmHg}}\right)\text{=}\frac{{\text{ΔH}}^{\text{o}}}{\text{8}\text{.314}\text{J/mol}\text{.K}}\left(\frac{\text{1}}{\text{323}\text{K}}\text{-}\frac{\text{1}}{\text{303}\text{K}}\right)\\ \\ \text{-1}\text{.067}\text{=}{\text{ΔH}}^{\text{o}}\left(\text{-2}\text{.458}{\text{×10}}^{\text{-5}}\right)\\ \\ {\text{ΔH}}^{\text{o}}\text{=}\text{4}\text{.34}{\text{×10}}^{\text{4}}\text{J/mol}\\ \\ \text{=}43.4kJ/mol\end{array}$

Therefore, the molar heat of vaporization of water is $43.4kJ/mol$